Categories: Physics, Plasma physics.

Lawson criterion

For sustained nuclear fusion to be possible, the Lawson criterion must be met, from which some required properties of the plasma and the reactor chamber can be deduced.

Suppose that a reactor generates a given power PfusP_\mathrm{fus} by nuclear fusion, but that it leaks energy at a rate PlossP_\mathrm{loss} in an unusable way. If an auxiliary input power PauxP_\mathrm{aux} sustains the fusion reaction, then the following inequality must be satisfied in order to have harvestable energy:

PlossPfus+Paux\begin{aligned} P_\mathrm{loss} \le P_\mathrm{fus} + P_\mathrm{aux} \end{aligned}

We can rewrite PauxP_\mathrm{aux} using the definition of the energy gain factor QQ, which is the ratio of the output and input powers of the fusion reaction:

QPfusPaux    Paux=PfusQ\begin{aligned} Q \equiv \frac{P_\mathrm{fus}}{P_\mathrm{aux}} \quad \implies \quad P_\mathrm{aux} = \frac{P_\mathrm{fus}}{Q} \end{aligned}

Returning to the inequality, we can thus rearrange its right-hand side as follows:

PlossPfus+PfusQ=Pfus(1+1Q)=Pfus(Q+1Q)\begin{aligned} P_\mathrm{loss} \le P_\mathrm{fus} + \frac{P_\mathrm{fus}}{Q} = P_\mathrm{fus} \Big( 1 + \frac{1}{Q} \Big) = P_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big) \end{aligned}

We assume that the plasma has equal species densities ni=nen_i = n_e, so its total density n=2nin = 2 n_i. Then PfusP_\mathrm{fus} is as follows, where fiif_{ii} is the frequency with which a given ion collides with other ions, and EfusE_\mathrm{fus} is the energy released by a single fusion reaction:

Pfus=fiiniEfus=(niσv)niEfus=n24σvEfus\begin{aligned} P_\mathrm{fus} = f_{ii} n_i E_\mathrm{fus} = \big( n_i \Expval{\sigma v} \big) n_i E_\mathrm{fus} = \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus} \end{aligned}

Where σv\Expval{\sigma v} is the mean product of the velocity vv and the collision cross-section σ\sigma.

Furthermore, assuming that both species have the same temperature Ti=Te=TT_i = T_e = T, the total energy density WW of the plasma is given by:

W=32kBTini+32kBTene=3kBTn\begin{aligned} W = \frac{3}{2} k_B T_i n_i + \frac{3}{2} k_B T_e n_e = 3 k_B T n \end{aligned}

Where kBk_B is Boltzmann‚Äôs constant. From this, we can define the confinement time τE\tau_E as the characteristic lifetime of energy in the reactor, before leakage. Therefore:

τEWPloss    Ploss=3nkBTτE\begin{aligned} \tau_E \equiv \frac{W}{P_\mathrm{loss}} \quad \implies \quad P_\mathrm{loss} = \frac{3 n k_B T}{\tau_E} \end{aligned}

Inserting these new expressions for PfusP_\mathrm{fus} and PlossP_\mathrm{loss} into the inequality, we arrive at:

3nkBTτEn24σvEfus(Q+1Q)\begin{aligned} \frac{3 n k_B T}{\tau_E} \le \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big) \end{aligned}

This can be rearranged to the form below, which is the original Lawson criterion:

nτEQQ+112kBTσvEfus\begin{aligned} n \tau_E \ge \frac{Q}{Q + 1} \frac{12 k_B T}{\Expval{\sigma v} E_\mathrm{fus}} \end{aligned}

However, it turns out that the highest fusion power density is reached when TT is at the minimum of T2/σvT^2 / \Expval{\sigma v}. Therefore, we multiply by TT to get the Lawson triple product:

nTτEQQ+112kBT2σvEfus\begin{aligned} \boxed{ n T \tau_E \ge \frac{Q}{Q + 1} \frac{12 k_B T^2}{\Expval{\sigma v} E_\mathrm{fus}} } \end{aligned}

For some reason, it is often assumed that the fusion is infinitely profitable QQ \to \infty, in which case the criterion reduces to:

nTτE12kBT2σvEfus\begin{aligned} n T \tau_E \ge \frac{12 k_B T^2}{\Expval{\sigma v} E_\mathrm{fus}} \end{aligned}


  1. M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.